what does the notation $x^{(n)}$ mean?












0












$begingroup$


What does the notation $x^{(n)}$ where $x$ is a matrix and $n$ is an integer?



$$|W^Tmathbf{x}^{(n)}+b-y^{(n)}|_2^2$$






optimization notation







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    0












    $begingroup$


    What does the notation $x^{(n)}$ where $x$ is a matrix and $n$ is an integer?



    $$|W^Tmathbf{x}^{(n)}+b-y^{(n)}|_2^2$$






    optimization notation







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      What does the notation $x^{(n)}$ where $x$ is a matrix and $n$ is an integer?



      $$|W^Tmathbf{x}^{(n)}+b-y^{(n)}|_2^2$$






      optimization notation







      share|cite|improve this question











      $endgroup$




      What does the notation $x^{(n)}$ where $x$ is a matrix and $n$ is an integer?



      $$|W^Tmathbf{x}^{(n)}+b-y^{(n)}|_2^2$$






      optimization notation




      optimization notation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 16 at 2:01









      David M.

      1,734418




      1,734418










      asked Jan 15 at 20:11









      Ibrahim AbouhashishIbrahim Abouhashish

      367




      367






















          2 Answers
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          3












          $begingroup$

          It is just an index for a sequence $x^{(1)}, x^{(2)}, x^{(3)}, ldots$, similar to $x_1, x_2, x_3, ldots$.



          The reason why they avoid subscripts $x_i$ is because presumably $x$ is a vector, so $x_i$ may be reserved for denoting "the $i$th component of the vector $x$." Using the above notation allows one to do things like $x^{(2)}_3$, which denotes the third component of the vector $x^{(2)}$ (which itself is the second element in a sequence of vectors).



          Finally, the reason for the parentheses is to differentiate it from an exponent, since "$y^2$" might look like the square of $y$.






          share|cite|improve this answer









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            3












            $begingroup$

            As I see the optimization tag, this probably means the matrix that is yielded at the $n$-th iteration step.



            In simple words, this means that $x^{(n)}$ is the $x$ matrix of your method at the $n$-th step and $y^{(n)}$ is the $y$ matrix of your method at the $n$-th step.






            share|cite|improve this answer











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              2 Answers
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              3












              $begingroup$

              It is just an index for a sequence $x^{(1)}, x^{(2)}, x^{(3)}, ldots$, similar to $x_1, x_2, x_3, ldots$.



              The reason why they avoid subscripts $x_i$ is because presumably $x$ is a vector, so $x_i$ may be reserved for denoting "the $i$th component of the vector $x$." Using the above notation allows one to do things like $x^{(2)}_3$, which denotes the third component of the vector $x^{(2)}$ (which itself is the second element in a sequence of vectors).



              Finally, the reason for the parentheses is to differentiate it from an exponent, since "$y^2$" might look like the square of $y$.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                It is just an index for a sequence $x^{(1)}, x^{(2)}, x^{(3)}, ldots$, similar to $x_1, x_2, x_3, ldots$.



                The reason why they avoid subscripts $x_i$ is because presumably $x$ is a vector, so $x_i$ may be reserved for denoting "the $i$th component of the vector $x$." Using the above notation allows one to do things like $x^{(2)}_3$, which denotes the third component of the vector $x^{(2)}$ (which itself is the second element in a sequence of vectors).



                Finally, the reason for the parentheses is to differentiate it from an exponent, since "$y^2$" might look like the square of $y$.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  It is just an index for a sequence $x^{(1)}, x^{(2)}, x^{(3)}, ldots$, similar to $x_1, x_2, x_3, ldots$.



                  The reason why they avoid subscripts $x_i$ is because presumably $x$ is a vector, so $x_i$ may be reserved for denoting "the $i$th component of the vector $x$." Using the above notation allows one to do things like $x^{(2)}_3$, which denotes the third component of the vector $x^{(2)}$ (which itself is the second element in a sequence of vectors).



                  Finally, the reason for the parentheses is to differentiate it from an exponent, since "$y^2$" might look like the square of $y$.






                  share|cite|improve this answer









                  $endgroup$



                  It is just an index for a sequence $x^{(1)}, x^{(2)}, x^{(3)}, ldots$, similar to $x_1, x_2, x_3, ldots$.



                  The reason why they avoid subscripts $x_i$ is because presumably $x$ is a vector, so $x_i$ may be reserved for denoting "the $i$th component of the vector $x$." Using the above notation allows one to do things like $x^{(2)}_3$, which denotes the third component of the vector $x^{(2)}$ (which itself is the second element in a sequence of vectors).



                  Finally, the reason for the parentheses is to differentiate it from an exponent, since "$y^2$" might look like the square of $y$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 15 at 20:17









                  angryavianangryavian

                  41.4k23381




                  41.4k23381























                      3












                      $begingroup$

                      As I see the optimization tag, this probably means the matrix that is yielded at the $n$-th iteration step.



                      In simple words, this means that $x^{(n)}$ is the $x$ matrix of your method at the $n$-th step and $y^{(n)}$ is the $y$ matrix of your method at the $n$-th step.






                      share|cite|improve this answer











                      $endgroup$


















                        3












                        $begingroup$

                        As I see the optimization tag, this probably means the matrix that is yielded at the $n$-th iteration step.



                        In simple words, this means that $x^{(n)}$ is the $x$ matrix of your method at the $n$-th step and $y^{(n)}$ is the $y$ matrix of your method at the $n$-th step.






                        share|cite|improve this answer











                        $endgroup$
















                          3












                          3








                          3





                          $begingroup$

                          As I see the optimization tag, this probably means the matrix that is yielded at the $n$-th iteration step.



                          In simple words, this means that $x^{(n)}$ is the $x$ matrix of your method at the $n$-th step and $y^{(n)}$ is the $y$ matrix of your method at the $n$-th step.






                          share|cite|improve this answer











                          $endgroup$



                          As I see the optimization tag, this probably means the matrix that is yielded at the $n$-th iteration step.



                          In simple words, this means that $x^{(n)}$ is the $x$ matrix of your method at the $n$-th step and $y^{(n)}$ is the $y$ matrix of your method at the $n$-th step.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Jan 15 at 20:46









                          Bernard

                          121k740116




                          121k740116










                          answered Jan 15 at 20:13









                          RebellosRebellos

                          14.8k31248




                          14.8k31248






























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